Pulleys with Torque: Free Body Diagram Analysis

[rashida564]

Homework Statement

In the diagram to the left, a mass m1 is connected through a fixed pulley to a second movable pulley attached to a mass m2, with a massless wire terminating at the ceiling. The two pulleys are identical, have mass M and radius T,and have moment of inertia 1/2MR^2 for rotation about their axes
1)Sketch free-body diagrams for this arrangement
2)What is m1/m2 when the two blocks are in equilibrium?
3)Write down the four equations describing the motion of the blocks. Hence, find an expression for the acceleration of the blocks in terms of the parameters of the problem and g.

Relevant Equations

F=ma
t=I(alpha)

Free body diagram is below, but I have confusion about the tension. The tension on the block and pulley should always be the same by Newton third law. so T1=T1' and T2=T2'.
Part B) The system is in equilibrium so net torque on each pulley is equal to zero. Therefore T1'=T2' and T2'=T3.
Applying Newton second law of motion 2T1'=m2g and T2'=m1g. So m1/m2=1/2.
Part C) Haven't attempted yet but I think a2 should be half of a1, then there's simultaneous equations to solve it
But my confusion is about T1 and T1'.

 

[Cutter Ketch]

Your diagram makes it very difficult for us (and probably you) to understand. I suppose we could figure it out from the text, but why not draw one diagram of the whole system so the relationships are clear?

 

[rashida564]

I hope this one is clearer

 

[Cutter Ketch]

Also, your free body diagrams have several errors that are causing confusion. Don’t forget the pulleys have mass, and also where is the connection between mass 2 and pulley 2?

 

[Cutter Ketch]

View attachment 255424
I hope this one is clearer

much clearer, thank you.

 

[rashida564]

Also, your free body diagrams have several errors that are causing confusion. Don’t forget the pulleys have mass, and also where is the connection between mass 2 and pulley 2?

I am thinking of not including the mass since I will only write the torque equation for the pulley

 

[Cutter Ketch]

I see nothing in the problem statement to suggest the mass of the pulleys can be ignored. It could be a reasonable approximation for light pulleys. However here they explicitly tell you they have mass. Besides, if they were light you could ignore the rotational inertia too, and that clearly isn’t the intention.

 

[rashida564]

I see nothing in the problem statement to suggest the mass of the pulleys can be ignored. It could be a reasonable approximation for light pulleys. However here they explicitly tell you they have mass. Besides, if they were light you could ignore the rotational inertia too, and that clearly isn’t the intention.

Thx I will have another go in it

 

[rashida564]

So I added the weight of the pulley in the diagram. And wrote the force equations. Where small t net torque. and anticlockwise is the positive direction.
for the pulley in the left t=(T1'-T2)R=1/2MR^2α1. Or T1'-T2=1/2Ma1
torque equation for the right pulley R(T2'-T3)=1/2MR^2α2. T2-T3=1/2Ma2
Force equation For block 1 m1g-T1=m1a
For equation for the system of Block 2 + Pulley " Ignoring the internal forces"
m2g+Mg-T2+T3=m2a

 

[Cutter Ketch]

You said, “for the pulley in the left
t=(T1'-T2)R=1/2MR^2α1. Or T1'-T2=1/2Ma1”

When relating the angular acceleration to the linear acceleration, why did both R’s disappear?

 

[rashida564]

One R Cancelled from the one in the left hand side and the other because α =a/R

 

[Cutter Ketch]

You said, “For equation for the system of Block 2 + Pulley " Ignoring the internal forces"
m2g+Mg-T2+T3=m2a”

Why is only the mass m2 accelerating?

 

[rashida564]

Sorry I forgot to include the other mass

 

[rashida564]

IS the other equation correct, I have the confusion about the direction of positive and negative with forces and torques

 

[haruspex]

2)What is m1/m2 when the two blocks are in equilibrium?

I must be missing something. Why isn't it the ratio of m₁ to m₂+M?

 

[rashida564]

This made me stop solving the question and assuming that my solution is wrong . It's taken from a past exam in my uni

 

[haruspex]

This made me stop solving the question and assuming that my solution is wrong . It's taken from a past exam in my uni

Ok, but we can continue with part C. What equations do you have for that now?

 

[Cutter Ketch]

One R Cancelled from the one in the left hand side and the other because α =a/R

Oops! My bad. Sorry!

 

[Cutter Ketch]

This made me stop solving the question and assuming that my solution is wrong . It's taken from a past exam in my uni

Oh no! You shouldn’t be discouraged. You have it. Your equations in post 9 are well constructed and show a good understanding of how to do the problem. If you left off a mass in one equation or didn’t draw one of the forces on pulley 2 in your force diagrams, that is only a little bit of attention to detail from perfect. Yes, as you mention in post 14 you now have to relate the signs correctly among the 4 equations, but that’s really the last step in setting up the problem.

Regarding relating the equations, what quantity connects them? What do they all share?

 

[haruspex]

Oh no! You shouldn’t be discouraged. You have it. Your equations in post 9 are well constructed and show a good understanding of how to do the problem. If you left off a mass in one equation or didn’t draw one of the forces on pulley 2 in your force diagrams, that is only a little bit of attention to detail from perfect. Yes, as you mention in post 14 you now have to relate the signs correctly among the 4 equations, but that’s really the last step in setting up the problem.

Regarding relating the equations, what quantity connects them? What do they all share?

I assume post #16 is in response to my post #15.

 

[rashida564]

All the equation contains 4 unknown with four equations, that relates the forces (after simplifying the torques equation) with the acceleration. Also I make the connected the acceleration of the a1 and a2 by the constrain equation. But I am scared of the negative and positive torque and how to define the positive direction

 

[haruspex]

the negative and positive torque and how to define the positive direction

This can be tricky. There are formal methods involving right hand rules and vector products, but it’s generally easier just to apply a little thought.
You don't need to guess correctly in advance which way things will move; just choose a positive direction for each force, torque and acceleration and be consistent. To connect forces with torques, is it clear to you that an upward force about an axis to the left leads to an anticlockwise torque?
For pulley problems like this, there is a cheat method. But I won't go into that if you can handle the method above.
Please post your equations.

 

[rashida564]

will the angular acceleration of both pulley be the same since they are connected by a rope

 

[haruspex]

will the angular acceleration of both pulley be the same since they are connected by a rope

I see no reason why they would.
Please post your equations.

 

[rashida564]

I asked my friend he told me both of them should have the same acceleration be the angular acceleration is tangential acceleration divided by the radius. Both of them have the same tangential acceleration "otherwise the rope will extend" and they have the same radius. So they will have the same angular acceleration

 

[haruspex]

otherwise the rope will extend

No, if they have different tangential accelerations (ignoring for the moment that they have the same radius) it only means the section of rope connecting them will change in length.

 

[rashida564]

So I alpha is a/R is where a is the acceleration of each pulley

 

[rashida564]

So far I got 4 equations of motions.
First for force assume up to be the positive direction
Constrain equation
a1=2a2
T1-m1g=m1a1=2m1a2
T2+T3-(M+m2)g=m2a2
For the torque equation, I assumed that counter clock wise is positive and also assumed that m1 goes down.
T1R-T2R=1/2 M R^2 (a2/R)(2). Since a1=2a2
so T1-T2=a2M
-T2R+T3R=1/2 MR^2 (-a2/R)
T2-T3=Ma2/2

 

[rashida564]

for part b) I putted a=0 So
T1-m1g=0
T2+T3-(M+m2)g=0
T1-T2=0
2T2-2T3=0
So T2=T3 for the last equation
Then T2=T1 for the third equation
So after that T1=mg from the first equation therefor T2=T3=T1=m1g
From the second equation
2m1g=M+m2
so m1=(M+m2)/2
Which kinda make sense, since the mechanical advantage is 2.
Now what's left is part c.

 

[haruspex]

So I alpha is a/R is where a is the acceleration of each pulley

Where a is the tangential acceleration in each case.

 

[haruspex]

a1=2a2
T1-m1g=m1a1=2m1a2
T2+T3-(M+m2)g=m2a2

Those equations are not consistent in regard to which ways are positive for the accelerations.

 

[rashida564]

I think I can see it now, is it because a1=-2a2

 

[haruspex]

I think I can see it now, is it because a1=-2a2

Yes.

 

[rashida564]

I think now I got it. By putting one of the linear equation as negative. namely : m1g-T1=2m1a2.
For here I added two equation T1-T2+(m1g-T1)=Ma2+m1a1
or m1g-T2=Ma2+2m1a2
For this I can find T2=m1g-Ma2-2m1a2
then I can also find T3, 2T2-2T3=Ma2
Or T3= (2T2-Ma2)/2
which is cool thing since T2 is expressed purely in terms of m, M,a and g.
Using the last equation T2+T3-(M+m2)g=m2a2
And substitute value of T2 and T3. Hopping I didn't make any mistake I will get that
a2=1/m2(((3m1-5m1a2)/2) + (M+m2)g )
And I can also find a1. which is just twice a2

 

[haruspex]

a2=1/m2(((3m1-5m1a2)/2) + (M+m2)g )

This is indecipherable. There appears to be a term (3m₁-5m₁a₂), which makes no sense. You can't subtract a force from an acceleration.
Please use parentheses as appropriate and either use LaTeX or use subscripts (use the ...▼ pulldown) and multiply out as necessary to avoid the divisions.
Also, for it to be an expression for a₂ you should arrange that a₂ does not occur on both sides of the equation.